If u, v, w are three non-coplanar unit vectors and α, β, γ are the angles between v and w, w and u,

Q: Wait, why is the bisector vector simply $\vec{v} + \vec{w}$? Shouldn't it be more complex?

That is a fantastic intuition! Usually, finding a bisector involves complex trigonometry. However, because $\vec{v}$ and $\vec{w}$ are unit vectors, they form the adjacent sides of a rhombus. In any rhombus, the diagonal bisects the angle between the sides. Since the diagonal is the vector sum $\vec{v} + \vec{w}$, it naturally points along the bisector. It is a beautiful geometric shortcut that saves us pages of calculation!

Q: How do I remember the identity $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2$?

Think of the scalar triple product as the volume of a parallelepiped. The cross products $\vec{a} \times \vec{b}$, $\vec{b} \times \vec{c}$, and $\vec{c} \times \vec{a}$ define a new parallelepiped. Through vector expansion, you can prove that the volume of this new shape is exactly the square of the volume of the original one. It is a standard result in JEE Advanced, so I highly recommend committing it to memory—it is a lifesaver in competitive exams!

Q: Why do we square the result at the end? Is it just to match the proof?

Not at all! We square it because the problem asks us to evaluate the box product of cross products: $[\vec{x} \times \vec{y} \quad \vec{y} \times \vec{z} \quad \vec{z} \times \vec{x}]$. As per the identity we discussed, this expression is inherently the square of the box product of the vectors $\vec{x}, \vec{y},$ and $\vec{z}$. We aren't just squaring it to make it look like the answer; we are squaring it because that is the fundamental definition of the expression given.

Question

If $\vec{u}, \vec{v}, \vec{w}$, are three non-coplanar unit vectors and $\alpha, \beta, \gamma$ are the angles between $\vec{v}$ and $\vec{w}$, $\vec{w}$ and $\vec{u}$, $\vec{u}$ and $\vec{v}$ respectively and $\vec{x}, \vec{y}, \vec{z}$ are unit vectors along the bisectors of the angles $\alpha, \beta, \gamma$ respectively. Prove that $[\vec{x} 	imes \vec{y} \quad \vec{y} 	imes \vec{z} \quad \vec{z} 	imes \vec{x}] = \frac{1}{16} [\vec{u} \quad \vec{v} \quad \vec{w}]^2 \sec^2 \frac{\alpha}{2} \sec^2 \frac{\beta}{2} \sec^2 \frac{\gamma}{2}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system with three non-coplanar unit vectors: $\vec{u}$, $\vec{v}$, and $\vec{w}$. Because they are unit vectors, any two of them, such as $\vec{v}$ and $\vec{w}$, form the sides of a rhombus.

In a rhombus, the diagonal acts as the angle bisector. The vector sum $\vec{v} + \vec{w}$ represents this diagonal. To find the unit bisector $\vec{x}$, we must normalize this sum.

We calculate the magnitude using the law of cosines for vectors:

$$|\vec{v} + \vec{w}| = \sqrt{|\vec{v}|^2 + |\vec{w}|^2 + 2|\vec{v}||\vec{w}|\cos\alpha}$$

Given $|\vec{v}| = |\vec{w}| = 1$, this simplifies to $\sqrt{2 + 2\cos\alpha}$. Applying the half-angle identity $1 + \cos\alpha = 2\cos^2(\alpha/2)$, the magnitude becomes $2\cos(\alpha/2)$.

Thus, the unit bisector $\vec{x}$ is defined as:

$$\vec{x} = \frac{1}{2} \sec\left(\frac{\alpha}{2}ight) (\vec{v} + \vec{w})$$

By symmetry, we can define the other unit bisectors $\vec{y}$ and $\vec{z}$ using the same logic. The beauty of this approach is that once one vector is defined, the others follow the same pattern.

### The Power of the Box Product

The problem requires us to evaluate the scalar triple product of cross products: $[\vec{x} 	imes \vec{y} \quad \vec{y} 	imes \vec{z} \quad \vec{z} 	imes \vec{x}]$. While this appears complex, we can utilize a fundamental identity in vector algebra:

$$[\vec{a} 	imes \vec{b} \quad \vec{b} 	imes \vec{c} \quad \vec{c} 	imes \vec{a}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2$$

This identity allows us to bypass the calculation of individual cross products. We simply need to determine the box product of the bisectors $\vec{x}, \vec{y},$ and $\vec{z}$, and then square the result.

### The Final Synthesis

Substituting our expressions for $\vec{x}, \vec{y},$ and $\vec{z}$ into the box product, we get:

$$[\vec{x} \quad \vec{y} \quad \vec{z}] = \left[ \frac{1}{2} \sec\frac{\alpha}{2} (\vec{v}+\vec{w}) \quad \frac{1}{2} \sec\frac{\beta}{2} (\vec{w}+\vec{u}) \quad \frac{1}{2} \sec\frac{\gamma}{2} (\vec{u}+\vec{v}) ight]$$

Using the linearity of the scalar triple product, we extract the constants:

$$\frac{1}{8} \sec\frac{\alpha}{2} \sec\frac{\beta}{2} \sec\frac{\gamma}{2} [\vec{v}+\vec{w} \quad \vec{w}+\vec{u} \quad \vec{u}+\vec{v}]$$

The inner box product of the sums of vectors is a known result:

$$[\vec{v}+\vec{w} \quad \vec{w}+\vec{u} \quad \vec{u}+\vec{v}] = 2[\vec{u} \quad \vec{v} \quad \vec{w}]$$

Combining these, the $2$ in the numerator cancels with the $8$ in the denominator to yield $1/4$. Squaring the entire expression as required, we arrive at the **final result**:

$$\frac{1}{16} [\vec{u} \quad \vec{v} \quad \vec{w}]^2 \sec^2 \left(\frac{\alpha}{2}ight) \sec^2 \left(\frac{\beta}{2}ight) \sec^2 \left(\frac{\gamma}{2}ight)$$

### Conclusion

You have successfully dismantled a daunting expression through geometric intuition and algebraic identities. This process demonstrates that complex systems often obey simple, elegant laws. Carry this confidence forward into your future challenges.

Visualized Solution

Visualizing the Vectors

Defining the Bisector Vector $x$

Magnitude of $v + w$

Calculating the Unit Vector $x$

Symmetry for $y$ and $z$

The Scalar Triple Product Property

Setting up $[x y z]$

Factoring Out Constants

Simplifying the Inner Box Product

Final Expression for $[x y z]$

The Final Square

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Power of the Box Product

The Final Synthesis

Conclusion

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Visualized Solution

Visualizing the Vectors

Defining the Bisector Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v+w

Calculating the Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Symmetry for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z